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Let $|G|=30$. I have prove that there is the only subgroup of order $15$, which I'll denote $H$. Now I do know how to classify the group. After thinking, I made the following steps.

1) Possible order of subgroup $K$ of $G$ of order 2 are 1, 3, 5, 15.

Case 1. if $G$ contain only one element of order 2, then $G \cong Z_{30}$.

Now I cannot solve for the next steps. Please give me any hints or any other method.

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  • $\begingroup$ @Anindya, you really must make more efforts to write correctly your question, and not only from a grammatical point of view. For example, in (1), what does "possible subgroup...of order 2 are 1,3,5,15" mean?! $\endgroup$ – DonAntonio Feb 26 '13 at 4:30
  • $\begingroup$ I would like to point you to math.stackexchange.com/q/569226/61691. That question has been marked as a duplicate of this one, but it is higher voted and it has the higher voted answers. $\endgroup$ – azimut Dec 10 '20 at 8:47
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Hints:

1) There is only one possible abelian group of order $\,30\,$

2) Any group $\,G\,$ of order 30 has a subgroup $\,H\,$ of order $\,15\,$, which is normal and abelian -- in fact, cyclic -- (why and why?), and thus $\,G\cong H\rtimes Q\,$ , for some subgroup $\,Q:=\langle\,q\,\rangle\,$ of order two.

Since $\,\operatorname{Aut}(H)\cong C_2\times C_4\,$ (why?) , there are at least four possible homomorphisms $\,Q\to\operatorname{Aut}(H)\,$ , all of them convolutions: (i) mapping $\,q\,$ to the generator of the factor $\,C_2\,$ , (ii) to $\,p^2\;,\;\;p=$ the generator of $\,C_4\,$ , and (iii) to the element $\,(q,p)\in C_2\times C_4\,$ (the trivial homomorphism gives the abelian group we already had before).

Check the above three non-trivial homomorphisms give three non-isomorphic groups of order $\;30\;$ .

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  • $\begingroup$ Does $Q$ is normal? $\endgroup$ – Andy Feb 26 '13 at 4:58
  • $\begingroup$ Not in general, @AnindyaGhatak . In fact, it is normal iff $\,G\,$ is a direct product, and since both $\,H,Q\,$ are abelian we'd be back in case (1). $\endgroup$ – DonAntonio Feb 26 '13 at 5:01
  • $\begingroup$ in (ii), mapping q to p won't give a homomorphism. $\endgroup$ – azimut Nov 16 '13 at 15:12
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    $\begingroup$ @DonAntonio: I'm sorry. I didn't realize that you wrote "Hints:" in the first line. $\endgroup$ – azimut Nov 16 '13 at 19:05
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    $\begingroup$ @RohanRajagopal Whenever we have a semidirect product as in my answer, there exisats a homomorphism from the right factor into the automorphism group of the left factor. This is the explanation why there's stuff about automorphism groups in my answer. $\endgroup$ – DonAntonio Nov 26 '18 at 13:18
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There is a general description of groups with cyclic Sylow subgroups:

Marshall Hall, The Theory of Groups - Theorem 9.4.3.

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  • $\begingroup$ Here is a reference to this result at wikipedia. $\endgroup$ – Ben Mar 11 '18 at 2:09

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